Math 152 - Lecture Notes # 3 - David Maslanka THE CATENARY A ...

Math 152 - Lecture Notes # 3 - David Maslanka
THE CATENARY
A derivation of the equation of a hanging cable. (Catenary is derived from the Latin word for chain).
Consider the section AP from the lowest point A to a general point P = ( x , y ) on the cable (see the figure below) and
imagine the rest of the cable to have been removed. The forces acting on the cable are:
i. H = horizontal tension pulling at A
ii. T = tangetial tension pulling at P
iii. W = δ s = weight of s feet of cable of density δ pounds per foot.
To be in equilibrium, the horizontal and vertical components of T must just balance H and W respectively. Thus,
T cos φ = H and T sin φ = W = δ s .
So
T sin φ
T cos φ = tan φ = δ s
H .
But since tan φ = dy
dx , we see that dy
dx = δ s
H .
Therefore, upon differentiating both sides with respect to x, we obtain the second order DE
( * )
d 2 y
dx 2 = δ H
ds
dx = δ
H
1 +
2
⎛ dy ⎞
⎜ ⎟
⎝ dx ⎠
.

Problem _______________________________________________________________________________________
Show that y = a cosh ( x a ) + C satisfies the differential equation ( * ) with a = H δ
by doing the following:
1. Substitute z = dy
dx
and
dz
dx = d2 y
dx 2 in ( * ) and obtain a first order DE with dependent variable z .
2. Solve the first order DE in z .
Hint: This involves verifying that:
Then note that
⌠
⎮
⌡
1
d
1 + z 2 z = ln ( z + 1 + z z ) + C .
ln ( z + 1 + z z ) = sinh -1 ( z ) , the inverse hyperbolic sine of z .
3. Replace z in terms of y and solve the resulting first order DE for y .